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hram777 [196]
3 years ago
7

What is the probability that you will get a 3-digit combination correct if no digits in the combination can be repeated?

Mathematics
2 answers:
Alex787 [66]3 years ago
7 0
If the digit of 0 is allowed, then:

1/(10*9*8)= 1/720

If it is not permitted, then:

1/(9*8*7)= 1/504

I hope this helps you!

Brainliest answer is always appreciated!
sergejj [24]3 years ago
7 0

Answer:

The probability that you will get a 3-digit combination correct is:

   1/720

Step-by-step explanation:

We are asked to find the probability that you will get a 3-digit combination correct if no digits in the combination can be repeated.

Now the favorable number of outcomes=1 (which is the correct combination)

Total number of outcomes=  10×9×8

(first place has 10 choices of digits from 0-9 ,second has 9 choices excluding the one at first place and third place has 8 choice)

                                           =720

Probability= number of favorable outcomes/total number of outcomes

                 =  1/720

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masha68 [24]

Let x,y be the dimensions of the rectangle. We know the equations for both area and perimeter:

A=xy=36

P=2(x+y)=36 \iff x+y=18

So, we have  the following system:

\begin{cases}xy=36\\x+y=18\end{cases}

From the second equation, we can deduce

y=18-x

Plug this in the first equation to get

xy=x(18-x)=-x^2+18=36

Refactor as

x^2-18x+36=0

And solve with the usual quadratic formula to get

x=9\pm3\sqrt{5}

Both solutions are feasible, because they're both positive.

If we chose the positive solution, we have

x=9+3\sqrt{5} \implies y=18-x=18-9-3\sqrt{5}=9-3\sqrt{5}

If we choose the negative solution, we have

x=9-3\sqrt{5} \implies y=18-x=18-9+3\sqrt{5}=9+3\sqrt{5}

So, we're just swapping the role of x and y. The two dimensions of the rectangle are 9+3\sqrt{5} and 9-3\sqrt{5}

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3 years ago
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Answer:

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